Converting FFT to inverse-FFT

On Tue, 15 Dec 2009 13:50:53 -0800 (PST), HardySpicer <gyansorova@gmail.com>
wrote:

>In fact the ordinary FFT should have a 1/N, not the inverse. Makes
>more sense. For example for DC you need to divide by N ie the average.

And yet, for non-DC terms, you need 2/N for the bin magnitude to equal the
sinewave amplitude.

Using 1/N is more mathematically consistent, though, because a real sine wave of
amplitude "A" will produce two spectral components, each of magnitude "A/2".

Greg

On Tue, 15 Dec 2009 13:50:53 -0800 (PST), HardySpicer <gyansorova@gmail.com>
wrote:

>In fact the ordinary FFT should have a 1/N, not the inverse. Makes
>more sense. For example for DC you need to divide by N ie the average.

And yet, for non-DC terms, you need 2/N for the bin magnitude to equal the
sinewave amplitude.

Using 1/N is more mathematically consistent, though, because a real sine wave of
amplitude "A" will produce two spectral components, each of magnitude "A/2".

Greg

On Tue, 15 Dec 2009 13:50:53 -0800 (PST), HardySpicer <gyansorova@gmail.com>
wrote:

>In fact the ordinary FFT should have a 1/N, not the inverse. Makes
>more sense. For example for DC you need to divide by N ie the average.

And yet, for non-DC terms, you need 2/N for the bin magnitude to equal the
sinewave amplitude.

Using 1/N is more mathematically consistent, though, because a real sine wave of
amplitude "A" will produce two spectral components, each of magnitude "A/2".

Greg